Mongolian Mathematical Olympiad

Answer: 19.
For $1 \le i \le 5$, let $a_i$ denote the $i$-th girl and let $A_i$ denote the set of students that $a_i$ knows. Similarly let $b_i$ denote the $i$-th boy and let $B_i$ denote the set of students that $b_i$ knows.
Since $|A_i \cap B_1| = i - 1$, $|A_i \cap B_5| = 5 - i$, we have
$$|A_1| \ge 4,\ |A_2| \ge 3,\ |A_3| \ge 2,\ |A_4| \ge 3,\ |A_5| \ge 4.$$
Similarly for $|B_i|$.
First suppose $|A_1| = 4$. Since $|A_1 \cap B_5| = 4$, we have $A_1 \subseteq B_5$, thus $A_1 \cap A_5 \subseteq B_5 \cap A_5 = \emptyset$. It follows that $|B_i| \ge |(A_1 \cup A_5) \cap B_i| = |A_1 \cap B_i| + |A_5 \cap B_i| = 4$. Hence $\sum |B_i| \ge 20$. Similarly, if $|A_5| = 4$, then $\sum |B_i| \ge 20$.
Now suppose $|A_3| = 2$. Then $|A_3 \cap B_1| = |A_3 \cap B_5| = 2$ implies that $A_3 \subseteq B_1 \cap B_5$. Hence $|B_1| \ge |A_5 \cap B_1| + |B_5 \cap B_1| \ge 6$. Analogously, we have $|B_5| \ge 6$, therefore $\sum |B_i| \ge 6 + 3 + 2 + 3 + 6 = 20$.
Finally, if $|A_1| \ge 5$, $|A_3| \ge 3$, $|A_5| \ge 5$, then we have $\sum |A_i| \ge 5 + 3 + 3 + 3 + 5 = 19$. Thus we have $S \ge 19$ and it suffices to find an example with $\sum |A_i| = \sum |B_i| = 19$:
$$\begin{align*} A_1 &= \{a_2, a_4, b_1, b_2, b_5\} & \{a_1, a_3, a_5, b_3, b_4\} &= B_1 \\ A_2 &= \{a_4, a_5, b_1\} & \{a_1, a_3, b_5\} &= B_2 \\ A_3 &= \{a_1, a_5, b_1\} & \{a_3, a_4, b_5\} &= B_3 \\ A_4 &= \{a_1, a_3, a_5\} & \{a_4, b_1, b_5\} &= B_4 \\ A_5 &= \{a_1, a_3, b_3, b_4, b_5\} & \{a_2, a_4, a_5, b_1, b_2\} &= B_5 \end{align*}$$